A block with mass m 16 kg rests on a frictionless table and is accelerated by a

Question

A block with mass m 16 kg rests on a frictionless table and is accelerated by a spring with spring constantk 5153 N/m after being compressed a distance X1 = 0.53 m from the spring's unstretched length. The floor is frictionless except for a rough patch a distance d = 2.9 m long. For this rough path, the coefficient of friction is -0.49 1) How much work is done by the spring as it accelerates the block? J Submit You currently have 0 submissions for this question. Only 9 submission are allowed. You can make 9 more submissions for this question. 2) What is the speed of the block right after it leaves the spring? m/s Submit You currently have 0 submissions for this question. Only 9 submission are allowed You can make 9 more submlsslons for this question. 3) How much work is done by friction as the block crosses the rough spot? J Submit You currently have 0 submisslons for this question. Only 9 submission are allowed. You can make 9 more submissions for this question. 4) What is the speed of the block after it passes the rough spot? m/s Submit You currently have 0 submissions for this question. Only 9 submission are allowed. You can make 9 more submissions for this question. 5) Instead, the spring is only compressed a distance x 0.118 m before being released. How far into the rough path does the block slide before coming to rest? m Submit You currently have 0 submissions for this question. Only 9 submission are allowed You can make 9 more submissions for this question. 6) What distance does the spring need to be compressed so that the block will just barely make it past the rough patch when released? m Submit You currently have 0 submissions for this question. Only 9 submission are allowed You can make 9 more submissions for this question. 7) If the spring was compressed three times farther and then the block is released, the work done on the block by the spring as it accelerates the block is: 0 the same three times greater O three times less O nine times greater O nine times less Submit You currently have 0 submissions for this question. Only 9 submission are allowed

Explanation / Answer

Given,

m = 16 kg ; k = 5153 N/m ; x1 = 0.53 m ; d = 2.9 m ; uk = 0.49

1)The work done by the spring will be:

Ws = 1/2 k x^2

Ws = 0.5 x 5153 x 0.53^2 = 723.74 J

Hence, Ws = 723.74 J

2)from conservation of energy

Ws = KE

1/2 m v^2 = 1/2 k x^2

v = x sqrt (k/m) = 0.53 x sqrt (5153/16) = 9.51 m/s

Hence, v = 9.51 m/s

3)work done by the friction will be:

Wf = uk m g d

wf = 0.49 x 16 x 9.81 x 2.9 = 223.04 J

Hence, Wf = 223.04 J

4)the energy of the block after rough patch will be:

E = Ws - Wf = 723.74 - 223.04 = 500.7

1/2 m v^2 = 500.7 => v = sqrt (2 x 500/16) = 7.91 m/s

Hence, 7.91 m/s

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